what could be said about the series of test-functions?

Question

in another thread an offtopic question came to my mind. Consider a test/bump function $\phi$, then consider all its derivatives $\phi^{(k)}$, of course they are bounded by constants $M_k$, next form the set $M_0, M_1, M_2,\ldots$ off all those bounds. As was pointed out in another thread this set itself may not be bounded, so heres my question. Is this enough to conclude that the series $$ \sum_{k=1}^{\infty} a_k \phi^{(k)}(0) \qquad a_k \in \mathbb{R} $$ is not in $\mathcal{D}'$ for $\phi \in \mathcal{D}$?

Answer 1

Unless your series converges you cannot say that it belongs to $\mathcal{D}'$. Suppose for instance that only finitely many $\phi^{(k)}(0)$'s are $0$. Then consider the sequence $a_k$ defined by $a_k=1/\phi^{(k)}(0)$ if $\phi^{(k)}(0) \ne 0$ and $a_k=0$ otherwise. We see that the series $\sum_{k=1}^\infty a_k\phi^{(k)}(0)$ diverges!

Answer 2

The series is not properly defined - do you mean $$\sum^{\infty}_{k=1}a_{k}M_{k}$$ instead? If you want to reason with $\mathcal{D}'$, then I guess you mean distributions. Yes, select $\phi=\sin[x]$, then $|\phi^{k}(x)|\le 1$ and we have have $\sum 1=\infty$. I guess you are suggesting that since $$M_{k}$$ is not bounded as $k\rightarrow \infty$, $a_{k}M_{k}$ could well blow up. This make sense since even for $\frac{1}{x}$ on an interval $[\epsilon,1]$, its derivatives's absolute value can be arbitrarily large.